3.10 \(\int \csc ^3(2 a+2 b x) \sin (a+b x) \, dx\)

Optimal. Leaf size=49 \[ -\frac{3 \csc (a+b x)}{16 b}+\frac{3 \tanh ^{-1}(\sin (a+b x))}{16 b}+\frac{\csc (a+b x) \sec ^2(a+b x)}{16 b} \]

[Out]

(3*ArcTanh[Sin[a + b*x]])/(16*b) - (3*Csc[a + b*x])/(16*b) + (Csc[a + b*x]*Sec[a + b*x]^2)/(16*b)

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Rubi [A]  time = 0.062518, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {4288, 2621, 288, 321, 207} \[ -\frac{3 \csc (a+b x)}{16 b}+\frac{3 \tanh ^{-1}(\sin (a+b x))}{16 b}+\frac{\csc (a+b x) \sec ^2(a+b x)}{16 b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[2*a + 2*b*x]^3*Sin[a + b*x],x]

[Out]

(3*ArcTanh[Sin[a + b*x]])/(16*b) - (3*Csc[a + b*x])/(16*b) + (Csc[a + b*x]*Sec[a + b*x]^2)/(16*b)

Rule 4288

Int[((f_.)*sin[(a_.) + (b_.)*(x_)])^(n_.)*sin[(c_.) + (d_.)*(x_)]^(p_.), x_Symbol] :> Dist[2^p/f^p, Int[Cos[a
+ b*x]^p*(f*Sin[a + b*x])^(n + p), x], x] /; FreeQ[{a, b, c, d, f, n}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2]
&& IntegerQ[p]

Rule 2621

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \csc ^3(2 a+2 b x) \sin (a+b x) \, dx &=\frac{1}{8} \int \csc ^2(a+b x) \sec ^3(a+b x) \, dx\\ &=-\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (-1+x^2\right )^2} \, dx,x,\csc (a+b x)\right )}{8 b}\\ &=\frac{\csc (a+b x) \sec ^2(a+b x)}{16 b}-\frac{3 \operatorname{Subst}\left (\int \frac{x^2}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{16 b}\\ &=-\frac{3 \csc (a+b x)}{16 b}+\frac{\csc (a+b x) \sec ^2(a+b x)}{16 b}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{16 b}\\ &=\frac{3 \tanh ^{-1}(\sin (a+b x))}{16 b}-\frac{3 \csc (a+b x)}{16 b}+\frac{\csc (a+b x) \sec ^2(a+b x)}{16 b}\\ \end{align*}

Mathematica [C]  time = 0.0166838, size = 29, normalized size = 0.59 \[ -\frac{\csc (a+b x) \text{Hypergeometric2F1}\left (-\frac{1}{2},2,\frac{1}{2},\sin ^2(a+b x)\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[2*a + 2*b*x]^3*Sin[a + b*x],x]

[Out]

-(Csc[a + b*x]*Hypergeometric2F1[-1/2, 2, 1/2, Sin[a + b*x]^2])/(8*b)

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Maple [A]  time = 0.037, size = 55, normalized size = 1.1 \begin{align*}{\frac{1}{16\,b\sin \left ( bx+a \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{2}}}-{\frac{3}{16\,b\sin \left ( bx+a \right ) }}+{\frac{3\,\ln \left ( \sec \left ( bx+a \right ) +\tan \left ( bx+a \right ) \right ) }{16\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(2*b*x+2*a)^3*sin(b*x+a),x)

[Out]

1/16/b/sin(b*x+a)/cos(b*x+a)^2-3/16/b/sin(b*x+a)+3/16/b*ln(sec(b*x+a)+tan(b*x+a))

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Maxima [B]  time = 1.88189, size = 1091, normalized size = 22.27 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(2*b*x+2*a)^3*sin(b*x+a),x, algorithm="maxima")

[Out]

1/32*(4*(3*sin(5*b*x + 5*a) + 2*sin(3*b*x + 3*a) + 3*sin(b*x + a))*cos(6*b*x + 6*a) - 12*(sin(4*b*x + 4*a) - s
in(2*b*x + 2*a))*cos(5*b*x + 5*a) + 4*(2*sin(3*b*x + 3*a) + 3*sin(b*x + a))*cos(4*b*x + 4*a) - 3*(2*(cos(4*b*x
+ 4*a) - cos(2*b*x + 2*a) - 1)*cos(6*b*x + 6*a) + cos(6*b*x + 6*a)^2 - 2*(cos(2*b*x + 2*a) + 1)*cos(4*b*x + 4
*a) + cos(4*b*x + 4*a)^2 + cos(2*b*x + 2*a)^2 + 2*(sin(4*b*x + 4*a) - sin(2*b*x + 2*a))*sin(6*b*x + 6*a) + sin
(6*b*x + 6*a)^2 + sin(4*b*x + 4*a)^2 - 2*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x
+ 2*a) + 1)*log((cos(b*x + 2*a)^2 + cos(a)^2 - 2*cos(a)*sin(b*x + 2*a) + sin(b*x + 2*a)^2 + 2*cos(b*x + 2*a)*s
in(a) + sin(a)^2)/(cos(b*x + 2*a)^2 + cos(a)^2 + 2*cos(a)*sin(b*x + 2*a) + sin(b*x + 2*a)^2 - 2*cos(b*x + 2*a)
*sin(a) + sin(a)^2)) - 4*(3*cos(5*b*x + 5*a) + 2*cos(3*b*x + 3*a) + 3*cos(b*x + a))*sin(6*b*x + 6*a) + 12*(cos
(4*b*x + 4*a) - cos(2*b*x + 2*a) - 1)*sin(5*b*x + 5*a) - 4*(2*cos(3*b*x + 3*a) + 3*cos(b*x + a))*sin(4*b*x + 4
*a) - 8*(cos(2*b*x + 2*a) + 1)*sin(3*b*x + 3*a) + 8*cos(3*b*x + 3*a)*sin(2*b*x + 2*a) + 12*cos(b*x + a)*sin(2*
b*x + 2*a) - 12*cos(2*b*x + 2*a)*sin(b*x + a) - 12*sin(b*x + a))/(b*cos(6*b*x + 6*a)^2 + b*cos(4*b*x + 4*a)^2
+ b*cos(2*b*x + 2*a)^2 + b*sin(6*b*x + 6*a)^2 + b*sin(4*b*x + 4*a)^2 - 2*b*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) +
b*sin(2*b*x + 2*a)^2 + 2*(b*cos(4*b*x + 4*a) - b*cos(2*b*x + 2*a) - b)*cos(6*b*x + 6*a) - 2*(b*cos(2*b*x + 2*
a) + b)*cos(4*b*x + 4*a) + 2*b*cos(2*b*x + 2*a) + 2*(b*sin(4*b*x + 4*a) - b*sin(2*b*x + 2*a))*sin(6*b*x + 6*a)
+ b)

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Fricas [A]  time = 0.500022, size = 230, normalized size = 4.69 \begin{align*} \frac{3 \, \cos \left (b x + a\right )^{2} \log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 3 \, \cos \left (b x + a\right )^{2} \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 6 \, \cos \left (b x + a\right )^{2} + 2}{32 \, b \cos \left (b x + a\right )^{2} \sin \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(2*b*x+2*a)^3*sin(b*x+a),x, algorithm="fricas")

[Out]

1/32*(3*cos(b*x + a)^2*log(sin(b*x + a) + 1)*sin(b*x + a) - 3*cos(b*x + a)^2*log(-sin(b*x + a) + 1)*sin(b*x +
a) - 6*cos(b*x + a)^2 + 2)/(b*cos(b*x + a)^2*sin(b*x + a))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(2*b*x+2*a)**3*sin(b*x+a),x)

[Out]

Timed out

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Giac [B]  time = 2.20544, size = 1918, normalized size = 39.14 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(2*b*x+2*a)^3*sin(b*x+a),x, algorithm="giac")

[Out]

-1/16*((tan(1/2*b*x + 2*a)*tan(1/2*a)^12 - 12*tan(1/2*b*x + 2*a)*tan(1/2*a)^10 + 6*tan(1/2*a)^11 - 27*tan(1/2*
b*x + 2*a)*tan(1/2*a)^8 - 2*tan(1/2*a)^9 - 36*tan(1/2*a)^7 + 27*tan(1/2*b*x + 2*a)*tan(1/2*a)^4 - 36*tan(1/2*a
)^5 + 12*tan(1/2*b*x + 2*a)*tan(1/2*a)^2 - 2*tan(1/2*a)^3 - tan(1/2*b*x + 2*a) + 6*tan(1/2*a))/((3*tan(1/2*b*x
+ 2*a)^2*tan(1/2*a)^5 - tan(1/2*b*x + 2*a)*tan(1/2*a)^6 - 10*tan(1/2*b*x + 2*a)^2*tan(1/2*a)^3 + 15*tan(1/2*b
*x + 2*a)*tan(1/2*a)^4 - 3*tan(1/2*a)^5 + 3*tan(1/2*b*x + 2*a)^2*tan(1/2*a) - 15*tan(1/2*b*x + 2*a)*tan(1/2*a)
^2 + 10*tan(1/2*a)^3 + tan(1/2*b*x + 2*a) - 3*tan(1/2*a))*(3*tan(1/2*a)^5 - 10*tan(1/2*a)^3 + 3*tan(1/2*a))) +
2*(tan(1/2*b*x + 2*a)^3*tan(1/2*a)^24 + 30*tan(1/2*b*x + 2*a)^3*tan(1/2*a)^22 - 6*tan(1/2*b*x + 2*a)^2*tan(1/
2*a)^23 + tan(1/2*b*x + 2*a)*tan(1/2*a)^24 - 756*tan(1/2*b*x + 2*a)^3*tan(1/2*a)^20 + 614*tan(1/2*b*x + 2*a)^2
*tan(1/2*a)^21 - 114*tan(1/2*b*x + 2*a)*tan(1/2*a)^22 + 6*tan(1/2*a)^23 + 2058*tan(1/2*b*x + 2*a)^3*tan(1/2*a)
^18 - 4578*tan(1/2*b*x + 2*a)^2*tan(1/2*a)^19 + 1932*tan(1/2*b*x + 2*a)*tan(1/2*a)^20 - 182*tan(1/2*a)^21 - 27
*tan(1/2*b*x + 2*a)^3*tan(1/2*a)^16 + 6210*tan(1/2*b*x + 2*a)^2*tan(1/2*a)^17 - 7462*tan(1/2*b*x + 2*a)*tan(1/
2*a)^18 + 1554*tan(1/2*a)^19 - 9396*tan(1/2*b*x + 2*a)^3*tan(1/2*a)^14 + 15588*tan(1/2*b*x + 2*a)^2*tan(1/2*a)
^15 - 2331*tan(1/2*b*x + 2*a)*tan(1/2*a)^16 - 2178*tan(1/2*a)^17 - 21924*tan(1/2*b*x + 2*a)^2*tan(1/2*a)^13 +
26028*tan(1/2*b*x + 2*a)*tan(1/2*a)^14 - 5668*tan(1/2*a)^15 + 9396*tan(1/2*b*x + 2*a)^3*tan(1/2*a)^10 - 21924*
tan(1/2*b*x + 2*a)^2*tan(1/2*a)^11 + 6468*tan(1/2*a)^13 + 27*tan(1/2*b*x + 2*a)^3*tan(1/2*a)^8 + 15588*tan(1/2
*b*x + 2*a)^2*tan(1/2*a)^9 - 26028*tan(1/2*b*x + 2*a)*tan(1/2*a)^10 + 6468*tan(1/2*a)^11 - 2058*tan(1/2*b*x +
2*a)^3*tan(1/2*a)^6 + 6210*tan(1/2*b*x + 2*a)^2*tan(1/2*a)^7 + 2331*tan(1/2*b*x + 2*a)*tan(1/2*a)^8 - 5668*tan
(1/2*a)^9 + 756*tan(1/2*b*x + 2*a)^3*tan(1/2*a)^4 - 4578*tan(1/2*b*x + 2*a)^2*tan(1/2*a)^5 + 7462*tan(1/2*b*x
+ 2*a)*tan(1/2*a)^6 - 2178*tan(1/2*a)^7 - 30*tan(1/2*b*x + 2*a)^3*tan(1/2*a)^2 + 614*tan(1/2*b*x + 2*a)^2*tan(
1/2*a)^3 - 1932*tan(1/2*b*x + 2*a)*tan(1/2*a)^4 + 1554*tan(1/2*a)^5 - tan(1/2*b*x + 2*a)^3 - 6*tan(1/2*b*x + 2
*a)^2*tan(1/2*a) + 114*tan(1/2*b*x + 2*a)*tan(1/2*a)^2 - 182*tan(1/2*a)^3 - tan(1/2*b*x + 2*a) + 6*tan(1/2*a))
/((tan(1/2*a)^12 - 30*tan(1/2*a)^10 + 255*tan(1/2*a)^8 - 452*tan(1/2*a)^6 + 255*tan(1/2*a)^4 - 30*tan(1/2*a)^2
+ 1)*(tan(1/2*b*x + 2*a)^2*tan(1/2*a)^6 - 15*tan(1/2*b*x + 2*a)^2*tan(1/2*a)^4 + 12*tan(1/2*b*x + 2*a)*tan(1/
2*a)^5 - tan(1/2*a)^6 + 15*tan(1/2*b*x + 2*a)^2*tan(1/2*a)^2 - 40*tan(1/2*b*x + 2*a)*tan(1/2*a)^3 + 15*tan(1/2
*a)^4 - tan(1/2*b*x + 2*a)^2 + 12*tan(1/2*b*x + 2*a)*tan(1/2*a) - 15*tan(1/2*a)^2 + 1)^2) - 3*log(abs(tan(1/2*
b*x + 2*a)*tan(1/2*a)^3 + 3*tan(1/2*b*x + 2*a)*tan(1/2*a)^2 - tan(1/2*a)^3 - 3*tan(1/2*b*x + 2*a)*tan(1/2*a) +
3*tan(1/2*a)^2 - tan(1/2*b*x + 2*a) + 3*tan(1/2*a) - 1)) + 3*log(abs(tan(1/2*b*x + 2*a)*tan(1/2*a)^3 - 3*tan(
1/2*b*x + 2*a)*tan(1/2*a)^2 + tan(1/2*a)^3 - 3*tan(1/2*b*x + 2*a)*tan(1/2*a) + 3*tan(1/2*a)^2 + tan(1/2*b*x +
2*a) - 3*tan(1/2*a) - 1)))/b